Proving $\sin\left(10^{\circ}\right)$ is Rational or Irrational

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The discussion centers on proving that $\sin(10^{\circ})$ is irrational. Utilizing the identity $\sin \alpha = 3\ \sin \frac{\alpha}{3} - 4\ \sin^{3} \frac{\alpha}{3}$, it is established that $\sin\left(10^{\circ}\right)$ corresponds to the root of the cubic equation $8\ x^{3} - 6\ x + 1 = 0$. The analysis shows that if $x$ were rational, it would lead to an impossible condition for integers $a$ and $b$, confirming that $\sin(10^{\circ})$ is indeed irrational.

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Is $\sin\left(10^{\circ}\right)$ rational or not? Prove it.
 
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Fallen Angel said:
Is $\sin\left(10^{\circ}\right)$ rational or not? Prove it.

[sp]From the relation...

$\displaystyle \sin \alpha = 3\ \sin \frac{\alpha}{3} - 4\ \sin^{3} \frac{\alpha}{3}\ (1)$

... it follows that $\displaystyle x= \sin \frac{\pi}{18}$ must be root of the cubic equation...

$\displaystyle 8\ x^{3} - 6\ x + 1 =0\ (2)$

If x is rational, then You can write $\displaystyle x = \frac{a}{b}$, being a and b integers... but in this case for (2) it must be...

$\displaystyle b = 4\ \sqrt{3\ a - \frac{1}{2}}\ (3)$

... and that's impossible for a and b integers... the conclusion is that x isn't rational...[/sp]

Kind regards

$\chi$ $\sigma$
 
Good work chisigma, my solution was almost the same.

From $8x^3-6x+1=0$, let $y=2 sin \ 10º$, then $y^3-3y+1=0$ and this polynomial has no rational roots (because of Gauss Lemma).
 

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